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I have to calcuate this limit: $$\lim _{x\to 0}\left(\frac{1-\frac{\left(1+x\right)^{\frac{1}{x}}}{e}}{x}\right)$$

I have no idea how to start, maybe taylor series?

Thanks.

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    $\begingroup$ Taylor series would work. You need to know that of $\ln(1+x)$ and $e^x$ around $0$: start with what is inside the parenthesis, remembering that $(1+x)^{1/x} = e^{\frac{1}{x}\ln(1+x)}$. Expansion to order 2 should be enough, at first glance. $\endgroup$
    – Clement C.
    Jan 26 '16 at 15:37
  • $\begingroup$ I have tried without success. Can you help me ? $\endgroup$ Jan 26 '16 at 15:39
  • $\begingroup$ See below. ${}{}{}$ $\endgroup$
    – Clement C.
    Jan 26 '16 at 15:45
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Detailed approach, using Taylor series. You will need the following two expansions (to order 2 and 1): $$\begin{align} \ln(1+x) &= x-\frac{x^2}{2} + o(x^2) \\ e^x &= 1+x + o(x) \end{align} $$ around $0$.

Now, starting from inside the parenthesis and working our way to the full expression: $$ \frac{1}{x}\ln(1+x) = 1-\frac{x}{2} + o(x) $$ so $$(1+x)^{1/x} = e^{\frac{1}{x}\ln(1+x)} = e^{1-\frac{x}{2} + o(x)} = e\cdot e^{-\frac{x}{2} + o(x)} = e(1-\frac{x}{2}+ o(x))$$ and $$ 1-\frac{(1+x)^{1/x} }{e} = 1 - (1-\frac{x}{2}+ o(x)) = \frac{x}{2}+ o(x). $$ Plugging it back for the coup de grâce, you get $$ \frac{1-\frac{(1+x)^{1/x} }{e}}{x} = \frac{1}{2}+ o(1) \xrightarrow[x\to0]{} \frac{1}{2}. $$

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Hint:

Substitute $y=\frac{1}{x}$, so your limit becomes $$ \lim_{y \to \infty}y\left(1-\frac{\left(1+\frac{1}{y} \right)^y}{e} \right) $$ now use the series expansion at $\infty$ $$ \left(1+\frac{1}{y}\right)^y=e-\frac{e}{2y} + O \left(\frac{1}{y^2}\right) $$ and you have the result .

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